   Chapter 8, Problem 73GQ

Chapter
Section
Textbook Problem

The equation for the combustion of gaseous methanol is2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) (a) Using the bond dissociation enthalpies in Table 8.8, estimate the enthalpy change for this reaction. What is the enthalpy of combustion of one mole of gaseous methanol? (b) Compare your answer in part (a) with the value of Δ t H ∘ calculated using enthalpies of formation data.

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction and the enthalpy of combustion of one mole of gaseous methanol has to be determined.

Concept Introduction:

Bond dissociation enthalpy is the energy required to break one mole of gaseous bonds to form gaseous atoms. The process of breaking bonds in a molecule is always endothermic therefore ΔrH for bond breaking is always positive and ΔrH value for formation of bond is always negative.

ΔrH=ΣΔH(bondsbrocken)ΣΔH(bondsformed)

Explanation

Equation for the given reaction is,

2CH3OH(g)+3O2(g)2CO2(g)+4H2O(g)

The enthalpy change for this reaction can be determined as follows,

Bonds Brocken:

6molofCH2molofOH2molofCO3molofO=O

ΣΔH(bondsbrocken)=6×413kJforCHbonds+2×463kJforOH+2×358kJforCObonds+3×498kJforO=O=5614

(b)

Interpretation Introduction

Interpretation:

Standard enthalpy change for the given reaction and the standard enthalpy of combustion of one mole of gaseous methanol has to be determined.

Concept Introduction:

Bond dissociation enthalpy is the energy required to break one mole of gaseous bonds to form gaseous atoms. The process of breaking bonds in a molecule is always endothermic therefore ΔrH for bond breaking is always positive and ΔrH value for formation of bond is always negative.

ΔrH=ΣΔH(bondsbrocken)ΣΔH(bondsformed)

Enthalpy of formation:

ΔrH=ΔfH0(products)ΔfH0(reactants)

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